It is not hard to see that, given one sample from a univariate unit-variance Gaussian $X\sim \mathcal{N}(\mu,1)$ with unknown $\mu$ s.t. $0<|\mu|\leq 1$, one can simulate one draw from a "Bernoulli" random variable $Y\in\{-1,1\}$ with $|\mathbb{E}[Y]| = \Theta(|\mu|)$. (Basically, just take $Y=\operatorname{sign}(X)$.)
Is it possible to do the converse (even allowing extra randomness, independent of the sample)? That is:
Given one realization of some random variable $Y\in\{-1,1\}$ with (unknown) $\mathbb{E}[Y] = \mu$, output $X\sim \mathcal{N}(\nu,1)$ s.t. $|\nu| = \Theta(|\mu|)$.
I conjecture that to be impossible (I have some mild contrived evidence), but I don't really see how one would go about proving that.